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  4. Let displaystyle∑n=0∞ (n3((2 n) !)+(2 n-1)(n !)/(n !)((2 n) !))=a e+(b/e)+c, where a, b, c ∈ Z and e= displaystyle∑n=0∞ (1/n !) Then a2-b+c is equal to

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Q. Let $\displaystyle\sum_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\frac{b}{e}+c$, where $a, b, c \in Z$ and $e=\displaystyle\sum_{n=0}^{\infty} \frac{1}{n !}$ Then $a^2-b+c$ is equal to ___

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Solution:

$ \displaystyle\sum_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)} $
$=\displaystyle\sum_{n=0}^{\infty} \frac{1}{(n-3) !}+\displaystyle\sum_{n=0}^{\infty} \frac{3}{(n-2) !} +\sum_{n=0}^{\infty} \frac{1}{(n-1) !}+\displaystyle\sum_{n=0}^{\infty} \frac{1}{(2 n-1) !}-\displaystyle\sum_{n=0}^{\infty} \frac{1}{(2 n) !} $
$ =e+3 e+e+\frac{1}{2}\left(e-\frac{1}{e}\right)-\frac{1}{2}\left(e+\frac{1}{e}\right)$
$ =5 e-\frac{1}{e} $
$ \therefore a^2-b+c=26$